Problem: $f\,^{\prime}(x)=6x^5+7$ and $f(-2)=30$. $f(1) = $
Answer: Finding $f(x)$ We have $f'(x)=6x^5+7$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ &= \int (6x^5+7)\,dx \\\\ & = {x^6+7x} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(-2)=30$. Here's what we get when we plug in $-2$ : $\begin{aligned}f(-2)&={(-2)^6+7(-2)} {+ C}\\\\ &={50} {+ C} \end{aligned}$ We are given that this must equal $30$ : $30 = {50} {+ C}$ Solving the equation gives us ${C=-20}$. Finding $f(1)$ Now, we have that $f(x)={x^6+7x} {-20}$. Let's find $f(1)$ by plugging in $1$ : $\begin{aligned}f(1)&=(1)^6+7(1) - 20\\\\ &=-12 \end{aligned}$ The answer $f(1) = -12$